3.46.87 \(\int \frac {e^{e^3+x} (64 x+32 x^2)}{64 e^{2 e^3+2 x} x^4-16 e^{e^3+x} x^2 \log (16)+\log ^2(16)} \, dx\)

Optimal. Leaf size=19 \[ \frac {4}{-8 e^{e^3+x} x^2+\log (16)} \]

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Rubi [A]  time = 0.64, antiderivative size = 21, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 4, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1593, 6688, 12, 6686} \begin {gather*} -\frac {4}{8 e^{x+e^3} x^2-\log (16)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(E^3 + x)*(64*x + 32*x^2))/(64*E^(2*E^3 + 2*x)*x^4 - 16*E^(E^3 + x)*x^2*Log[16] + Log[16]^2),x]

[Out]

-4/(8*E^(E^3 + x)*x^2 - Log[16])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{e^3+x} x (64+32 x)}{64 e^{2 e^3+2 x} x^4-16 e^{e^3+x} x^2 \log (16)+\log ^2(16)} \, dx\\ &=\int \frac {32 e^{e^3+x} x (2+x)}{\left (8 e^{e^3+x} x^2-\log (16)\right )^2} \, dx\\ &=32 \int \frac {e^{e^3+x} x (2+x)}{\left (8 e^{e^3+x} x^2-\log (16)\right )^2} \, dx\\ &=-\frac {4}{8 e^{e^3+x} x^2-\log (16)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 19, normalized size = 1.00 \begin {gather*} \frac {4}{-8 e^{e^3+x} x^2+\log (16)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^3 + x)*(64*x + 32*x^2))/(64*E^(2*E^3 + 2*x)*x^4 - 16*E^(E^3 + x)*x^2*Log[16] + Log[16]^2),x]

[Out]

4/(-8*E^(E^3 + x)*x^2 + Log[16])

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fricas [A]  time = 1.00, size = 19, normalized size = 1.00 \begin {gather*} -\frac {1}{2 \, x^{2} e^{\left (x + e^{3}\right )} - \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x^2+64*x)*exp(exp(3)+x)/(64*x^4*exp(exp(3)+x)^2-64*x^2*log(2)*exp(exp(3)+x)+16*log(2)^2),x, algo
rithm="fricas")

[Out]

-1/(2*x^2*e^(x + e^3) - log(2))

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giac [A]  time = 0.13, size = 19, normalized size = 1.00 \begin {gather*} -\frac {1}{2 \, x^{2} e^{\left (x + e^{3}\right )} - \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x^2+64*x)*exp(exp(3)+x)/(64*x^4*exp(exp(3)+x)^2-64*x^2*log(2)*exp(exp(3)+x)+16*log(2)^2),x, algo
rithm="giac")

[Out]

-1/(2*x^2*e^(x + e^3) - log(2))

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maple [A]  time = 0.35, size = 16, normalized size = 0.84

method	result	size
norman	\(\frac {1}{-2 x^{2} {\mathrm e}^{{\mathrm e}^{3}+x}+\ln \relax (2)}\)	\(16\)
risch	\(\frac {1}{-2 x^{2} {\mathrm e}^{{\mathrm e}^{3}+x}+\ln \relax (2)}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*x^2+64*x)*exp(exp(3)+x)/(64*x^4*exp(exp(3)+x)^2-64*x^2*ln(2)*exp(exp(3)+x)+16*ln(2)^2),x,method=_RETUR
NVERBOSE)

[Out]

1/(-2*x^2*exp(exp(3)+x)+ln(2))

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maxima [A]  time = 0.51, size = 19, normalized size = 1.00 \begin {gather*} -\frac {1}{2 \, x^{2} e^{\left (x + e^{3}\right )} - \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x^2+64*x)*exp(exp(3)+x)/(64*x^4*exp(exp(3)+x)^2-64*x^2*log(2)*exp(exp(3)+x)+16*log(2)^2),x, algo
rithm="maxima")

[Out]

-1/(2*x^2*e^(x + e^3) - log(2))

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mupad [B]  time = 3.30, size = 15, normalized size = 0.79 \begin {gather*} \frac {1}{\ln \relax (2)-2\,x^2\,{\mathrm {e}}^{{\mathrm {e}}^3}\,{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + exp(3))*(64*x + 32*x^2))/(64*x^4*exp(2*x + 2*exp(3)) + 16*log(2)^2 - 64*x^2*exp(x + exp(3))*log(2
)),x)

[Out]

1/(log(2) - 2*x^2*exp(exp(3))*exp(x))

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sympy [A]  time = 0.12, size = 17, normalized size = 0.89 \begin {gather*} - \frac {1}{2 x^{2} e^{x + e^{3}} - \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x**2+64*x)*exp(exp(3)+x)/(64*x**4*exp(exp(3)+x)**2-64*x**2*ln(2)*exp(exp(3)+x)+16*ln(2)**2),x)

[Out]

-1/(2*x**2*exp(x + exp(3)) - log(2))

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